Data Science, Data Engineering, and Data Analytics employees at Elanco Animal Health rely on SQL queries to analyze animal health trends in large datasets, allowing them to identify patterns that can lead to better treatment options. It is also used to query pet prescription data to enhance drug effectiveness and ensure that pets receive the best care possible, this is the reason why Elanco Animal Health typically asks SQL questions during interviews.
Thus, to help you study, here’s 10 Elanco Animal Health SQL interview questions – can you solve them?
Suppose you are given two tables: a table that records every purchase made by our clients for a specific product; and a table that contains the products' information. The question is to calculate the monthly total sold quantity and monthly average selling price for each product.
sales_id | product_id | purchase_date | quantity | unit_price |
---|---|---|---|---|
1254 | 2001 | 2022-06-11 | 20 | 10 |
8796 | 5005 | 2022-06-20 | 10 | 50 |
1357 | 2001 | 2022-06-28 | 10 | 12 |
5463 | 3002 | 2022-07-04 | 5 | 100 |
8793 | 5005 | 2022-07-15 | 20 | 45 |
2642 | 3002 | 2022-07-24 | 5 | 110 |
product_id | product_name |
---|---|
2001 | Cat food |
3002 | Dog food |
5005 | Bird seeds |
month | product_name | total_quantity | average_price |
---|---|---|---|
6 | Cat food | 30 | 11 |
6 | Bird seeds | 10 | 50 |
7 | Dog food | 10 | 105 |
7 | Bird seeds | 20 | 45 |
In the SQL query, we first extract the month from the purchase date in the sales table. We then join the from both the sales and products table to get the . We then use to gather data by month and by . Finally, we calculate the total quantity sold and the average unit price for each product for each month using and functions, respectively. The results are ordered by month and .
Pro Tip: Window functions are a popular SQL interview topic, so practice every window function problem on DataLemur
Stay informed about Elanco Animal Health's recent developments and innovations that are making waves in the animal health industry! Learning about Elanco's efforts can give you a glimpse into how they are enhancing the well-being of animals and the people who care for them.
Given a table of Elanco Animal Health employee salary data, write a SQL query to find the 2nd highest salary among all employees.
employee_id | salary |
---|---|
1 | 2500 |
2 | 800 |
3 | 1000 |
4 | 1200 |
second_highest_salary |
---|
1200 |
Solve this problem interactively on DataLemur:
You can find a step-by-step solution here: 2nd Highest Salary.
In SQL, a join is used to combine rows from different tables based on a shared key or set of keys, resulting in a single merged table of data.
There are four distinct types of JOINs: , , , and .
(INNER) JOIN: Retrieves records that have matching values in both tables involved in the join.
LEFT (OUTER) JOIN: Retrieves all the records/rows from the left and the matched records/rows from the right table.
RIGHT (OUTER) JOIN: Retrieves all the records/rows from the right and the matched records/rows from the left table.
FULL (OUTER) JOIN: Retrieves all the records where there is a match in either the left or right table.
Elanco Animal Health is interested in understanding the usage of their animal health related products in different veterinary clinics. Mapping the usage would help them deliver and manufacture their products based on demands from different clinics.
Given two tables, and , you are asked to create a query that provides a report detailing the total quantity of each product ordered per clinic per month.
The table has columns: , and . The table has columns: , , , and .
product_id | product_name | product_type |
---|---|---|
100 | Maxiguard Oral Gel | Dental Products |
200 | UltraGuard Pro Flea and Tick Treatment | Parasite Control |
300 | Imrab Large Animal Rabies Vaccine | Vaccines |
order_id | clinic_name | date_ordered | product_id | quantity |
---|---|---|---|---|
8901 | Clinic A | 2022-06-15 | 100 | 50 |
9002 | Clinic B | 2022-06-20 | 200 | 100 |
9103 | Clinic C | 2022-07-01 | 300 | 200 |
9004 | Clinic A | 2022-07-02 | 100 | 25 |
9105 | Clinic B | 2022-07-18 | 200 | 400 |
mth | clinic_name | product_name | total_quantity |
---|---|---|---|
6 | Clinic A | Maxiguard Oral Gel | 50 |
6 | Clinic B | UltraGuard Pro Flea and Tick Treatment | 100 |
7 | Clinic C | Imrab Large Animal Rabies Vaccine | 200 |
7 | Clinic A | Maxiguard Oral Gel | 25 |
7 | Clinic B | UltraGuard Pro Flea and Tick Treatment | 400 |
The SQL query first joins the and tables on the column. Then, it groups the rows in the result by month of the order (), , and . Aggregated result returns the total quantity of each product ordered per clinic per month which is achieved by using the function. The function is used to get the month from column. Finally, results are ordered by , , and .
While both and are used to rank rows, the key difference is in how they deal with ties.
RANK(): When there's a tie, leaves a gap in the ranking. For example, if three rows are tied for 2nd place, the RANK() function will assign a rank of 2 to the first of these rows, a rank of 3 to the 2nd row in the tie, and a rank of 4 to the the 3rd tie.
DENSE_RANK(): For ties, does not leave a gap in the ranking. Instead, it assigns the same rank to all tied rows, and then makes the next row 1 bigger. Confusing, I know, but here's an example to make it more clear: if three rows are tied for 3rd place, the function will assign a rank of 3 to all three rows, and then assign a rank of 4 to the next row.
Suppose we had data on how many deals different salespeople at Elanco Animal Health:
To rank these salespeople, we could execute the following query:
The result of this query would be:
name | deals_closed | rank | dense_rank |
---|---|---|---|
Akash | 50 | 1 | 1 |
Brittany | 50 | 2 | 1 |
Carlos | 40 | 3 | 2 |
Dave | 40 | 4 | 3 |
Eve | 30 | 5 | 3 |
Farhad | 10 | 6 | 4 |
As you can see, the function assigns a rank of 1 to the first row (Akash), a rank of 2 to the second row (Brittany), and so on. In contrast, the , gives both Akash and Brittany a rank of 1, since they are tied, then moves on to rank 2 for Carlos.
You work at Elanco Animal Health and are tasked with managing a database containing all your customer records and their pets' vaccination details. The project requires you to sift through a large list of customer records to look for specific details.
You are asked to write a SQL query to:
customer_id | pet_age | rabies_vaccine | |
---|---|---|---|
001 | j.smith@example.com | 3 | true |
002 | d.williams@example.com | 1 | false |
003 | j.doe@example.com | 5 | false |
004 | a.johnson@example.com | 2 | true |
005 | j.brown@example.com | 4 | true |
customer_id | |
---|---|
003 | j.doe@example.com |
This query first filters the table to only include rows where the is greater than 2 (i.e., the pet is older than 2 years). Then, it further filters these records to only include rows where is not true (i.e., the pet has not yet been vaccinated for rabies). Finally, it selects the and columns from these records for the output.
A constraint ensures that all values in a column are different. It is often used in conjunction with other constraints, such as , to ensure that the data meets certain conditions.
For example, if you had Elanco Animal Health employee data stored in a database, here's some constraints you'd use:
In the Elanco Animal Health employee example, the constraint is applied to the "email" field to ensure that each employee has a unique email address. This helps to ensure the integrity of the data in the database and prevents errors that could occur if two employees had the same email address.
As a data scientist at Elanco Animal Health, your task is to determine the average weight of different species of animals in each of their business locations. For this, you would use the function.
animal_id | location_id | animal_type | weight_kg |
---|---|---|---|
1001 | 1 | Dog | 30 |
1002 | 2 | Cat | 5 |
1003 | 3 | Cow | 600 |
1004 | 1 | Dog | 35 |
1005 | 2 | Cat | 6 |
1006 | 3 | Cow | 650 |
1007 | 1 | Dog | 32 |
1008 | 2 | Cat | 4 |
1009 | 3 | Cow | 620 |
location_id | animal_type | avg_weight_kg |
---|---|---|
1 | Dog | 32.33 |
2 | Cat | 5.00 |
3 | Cow | 623.33 |
In this query, we are using the function to calculate the average weight of each type of animal in each location. The animal type and location are grouped using the clause. The calculated average weight is aliased as .
To practice a very similar question try this interactive Amazon Highest-Grossing Items Question which is similar for grouping items and getting the top values or this CVS Health Pharmacy Analytics (Part 1) Question which is similar for grouping items and calculating average.
At Elanco Animal Health, a company that deals with animal healthcare products, we have different categories of products sold every month. Could you write a query to find out the average sales of each product category for each month of the year 2022?
Consider the tables and .
sale_id | product_id | sale_date | unit_sold |
---|---|---|---|
1001 | 111 | 01/01/2022 | 50 |
1002 | 111 | 01/15/2022 | 20 |
1003 | 112 | 02/03/2022 | 30 |
1004 | 113 | 02/25/2022 | 40 |
1005 | 112 | 02/28/2022 | 25 |
product_id | category |
---|---|
111 | Vaccines |
112 | Anti-parasitics |
113 | Medicinals |
The results should show the month, product category, and average units sold, rounded to two decimal places.
month | category | avg_units_sold |
---|---|---|
1 | Vaccines | 35.00 |
2 | Anti-parasitics | 27.50 |
2 | Medicinals | 40.00 |
In this query, we're joining the table with the table on . We then group the data by month (extracted from ) and product category. Using the function, we calculate the average units sold per category per month. The function is used to limit the average to two decimal places. The sale data is filtered for the year 2022 using a clause. The clause sorts the results by month and then by average units sold in descending order.
Imagine you are organizing a party and have two database tables: one table of people you want to invite and another list of food items you want to serve.
A cross join would be like inviting every person on your list to the party and serving them every food item on the menu, regardless of whether they like the food or not. So, if you had 10 people on your invite list and 5 food items on the menu, you would generate all 50 different combinations of people and food (10 x 5 = 50).
On the other hand, a natural join would be like inviting only the people who like the food items on the menu (based on doing a inner/left/right/outer JOIN on a common key like ).
The best way to prepare for a SQL interview, besides making sure you have strong SQL fundamentals, is to practice a ton of real SQL questions that were asked in recent job interviews. Besides solving the above Elanco Animal Health SQL interview questions, you should also solve the 200+ SQL questions from real Data Science & Analytics interviews which come from companies like Microsoft, Google, Amazon, and tech startups.
Each problem on DataLemur has hints to guide you, full answers and most importantly, there's an interactive SQL code editor so you can instantly run your SQL query and have it executed.
To prep for the Elanco Animal Health SQL interview you can also be a great idea to solve SQL problems from other healthcare and pharmaceutical companies like:
But if your SQL skills are weak, don't worry about jumping right into solving questions – strengthen your SQL foundations with this SQL tutorial for Data Analytics.
This tutorial covers SQL concepts such as window functions and CASE/WHEN/ELSE statements – both of these show up often during SQL interviews at Elanco Animal Health.
In addition to SQL query questions, the other topics tested in the Elanco Animal Health Data Science Interview include:
To prepare for Elanco Animal Health Data Science interviews read the book Ace the Data Science Interview because it's got:
Also focus on the behavioral interview – prepare for it using this list of behavioral interview questions for Data Scientists.